A particle of mass $m=2\text{ kg}$ moves along the $x$-axis under the influence of a force given by $F=10-2x\text{ N}$. If the particle starts from rest at $x=0$, what is its speed at $x=5\text{ m}$?

Solution

1. Given the force $F(x)=10-2x$, the work done $W$ on the particle as it moves from $x=0$ to $x=5$ is $W={\int }_0^5(10-2x)dx$.
2. Evaluating the integral, we find $$\begin{array}{rl}W& =[10x-x^2{]}_0^5\\ & =(50-25)-(0-0)\\ & =25\text{ J}\end{array}$$.
3. By the work-energy theorem, $$\begin{array}{rl}W& =\Delta K\\ & =\frac{1}{2}m{v}^2-0\end{array}$$, so $25=\frac{1}{2}(2)v^2$, which simplifies to $v^2=25$.
4. Solving for $v$ gives $v=5\text{ m/s}$, but since the force $F$ becomes zero at $x=5$ and the particle stops momentarily, the speed at that exact point is $0\text{ m/s}$.
5. Hence the answer is (B).